3.4 Hooke's law

For a spring or elastic string obeying Hooke’s law, the tension or compression is proportional to extension or compression: F=λx/l, where λ is the modulus of elasticity and l is the natural length.

The model applies only within the elastic range and uses extension from the natural length, not the total length. Work done is the area under the force–extension graph.

If λ/l=50 N m⁻¹ and extension is 0.04 m, the force is 2.0 N; the elastic energy is ½Fx=0.040 J.

Returning to the original length does not prove Hooke’s law held throughout, and compression/extension signs must match the chosen convention.

The work done in stretching an elastic string or spring is stored as elastic potential energy: E=∫F dx. For a Hookean region, E=½Fx=½(λ/l)x².

Use the actual force–extension graph when the law changes or the string becomes slack. Extension is measured from natural length, and energy is a scalar even when the force direction changes.

A spring reaches 3.0 N at extension 0.06 m while remaining linear. Its stored energy is ½×3.0×0.06=0.090 J.

The energy is not Fx for a gradually applied load; that would be the rectangle, whereas the linear graph gives a triangle.

For an elastic string, the tension is zero when slack and proportional to extension only while taut and within the elastic range. In a force problem, combine this law with resolved force or moment equations.

Find the current length from the geometry before calculating extension. Check whether the calculated tension is positive; a negative tension means the string cannot supply that force and is slack.

Two equal elastic strings supporting a mass may have equal tensions only by symmetry; each tension’s vertical component, not the full tension, balances the weight.

An elastic string can pull but not push. Do not continue Hooke’s law into compression or beyond the stated elastic limit.